NTA Abhyas JEE Main2020MathematicsQuadratic EquationPractice
The values of λ for which one root of the equation x 2 + 1 - 2 λ x + λ 2 - λ - 2 = 0 is greater than 3 and the other smaller than 2 are given by
Options
- A2 < λ < 5
- B1 < λ < 4
- C1 < λ < 5
- D2 < λ < 4
Correct answer
D. 2 < λ < 4
Step-by-step solution
D > 0 ,   f 2 < 0 and f 3 < 0 D = 1 - 2 λ 2 - 4 λ 2 - λ - 2 = 1 + 4 λ 2 - 4 λ - 4 λ 2 + 4 λ + 8 = 9 > 0 (always true) f 2 < 0 ⇒ 4 + 2 1 - 2 λ + λ 2 - λ - 2 < 0 ⇒ 4 + 2 - 4 λ + λ 2 - λ - 2 < - 0 ⇒ λ 2 - 5 λ + 4 < 0 ⇒ λ ∈ 1,4 ...(i) f 3 < 0 ⇒ 9 + 3 1 - 2 λ + λ 2 - λ - 2 < 0 ⇒ 9 + 3 - 6 λ + λ 2 - λ - 2 < 0 ⇒ λ